# I What is the outer product of a tensor product of vectors?

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1. Apr 26, 2016

### Frank Castle

If one has two single-particle Hilbert spaces $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$, such that their tensor product $\mathcal{H}_{1}\otimes\mathcal{H}_{2}$ yields a two-particle Hilbert space in which the state vectors are defined as $$\lvert\psi ,\phi\rangle =\lvert\psi\rangle\otimes\lvert\phi\rangle\in\mathcal{H}_{1}\otimes\mathcal{H}_{2}$$ where $\lvert\psi\rangle\in\mathcal{H}_{1}$ and $\lvert\phi\rangle\in\mathcal{H}_{2}$.

Now, the inner product for $\mathcal{H}_{1}\otimes\mathcal{H}_{2}$ is defined such that $$\langle\phi ,\psi\vert\psi ,\phi\rangle =\left(\langle\phi\rvert\otimes\langle\psi\lvert\right)\left(\lvert\psi\rangle\otimes\lvert\phi\rangle\right) =\langle\psi\lvert\psi\rangle_{1}\langle\phi\lvert\phi\rangle_{2}$$ where $\langle\cdot\lvert\cdot\rangle_{1}$ is the inner product defined on $\mathcal{H}_{1}$ and $\langle\cdot\lvert\cdot\rangle_{2}$ the inner product defined on $\mathcal{H}_{2}$.

How though is the outer product defined? Is it simply $$\lvert\psi ,\phi\rangle\langle\phi ,\psi\rvert =\lvert\psi\rangle\otimes\lvert\phi\rangle\langle\phi\rvert\otimes\langle\psi\rvert =\lvert\psi\rangle\langle\psi\rvert_{1} \lvert\phi\rangle\langle\phi\rvert_{2}$$ where $\lvert\psi\rangle\langle\psi\rvert_{1}$ is the outer product in $\mathcal{H}_{1}$ and $\lvert\phi\rangle\langle\phi\rvert_{2}$ is the outer product in $\mathcal{H}_{2}$.

2. Apr 26, 2016

### andrewkirk

It might be easier to think of more general outer products, rather than the outer product of an element with itself, as you have written.
That is, consider the outer product $|a,b\rangle\langle c,d|=|a\rangle\otimes |b\rangle\langle c|\otimes\langle d|$.

This is an operator on the product space $\mathcal{H}_{1}\otimes\mathcal{H}_{2}$. To see what it does, we apply it to an element of that space $|e,f\rangle=|e\rangle\otimes |f\rangle$.

The result is $|a,b\rangle \langle c|e\rangle \langle d|f\rangle$

3. Apr 26, 2016

### Frank Castle

So one can't define it as $\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\lvert b\rangle\langle d\rvert$ then? Should it be something like $$\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\otimes\lvert b\rangle\langle d\rvert$$

I ask in particular as I'm trying to understand the notion of a partial trace.

Last edited: Apr 26, 2016
4. Apr 26, 2016

### andrewkirk

That symbol string on the RHS is not intrinsically meaningful. It juxtaposes the two operators $\lvert a\rangle\langle c\rvert$ and $\lvert b\rangle\langle d\rvert$ but:
- since they are not scalars, the juxtaposition cannot be interpreted as scalar multiplication
- since they are operators on different spaces ($\mathcal H_1$ and $\mathcal H_2$), the juxtaposition cannot be interpreted as composition of operation.

That is one of a number of intrinsically meaningful ways of writing it. It has the same meaning as what I wrote above $( |a\rangle\otimes |b\rangle)(\langle c|\otimes\langle d|)$ (I have added parentheses here that were only implied above, to make it clear what operations are being performed).

5. Apr 27, 2016

### Frank Castle

How does one show that the two expressions are equivalent? Would it be something like this:
$$\left(\lvert a\rangle\otimes\lvert b\rangle\langle c\rvert\otimes\langle d\rvert\right)\lvert e\rangle\otimes\lvert f\rangle =\lvert a\rangle\otimes\lvert b\rangle\left(\langle c\vert e\rangle\langle d\vert f\rangle\right)\\ =\left(\lvert a\rangle\langle c\vert e\rangle\right)\otimes \left(\lvert b\rangle\langle d\vert f\rangle\right) =\left(\lvert a\rangle\langle c\lvert\otimes\lvert b\rangle\langle d\lvert\right)\lvert e\rangle\otimes\lvert f\rangle$$

6. Apr 27, 2016

### andrewkirk

I would write it as follows:

$$(|a,b\rangle\langle c,d|)|e,f\rangle\equiv (|a\rangle\otimes|b\rangle)(\langle c|\otimes\langle d|)(|e\rangle\otimes|f\rangle)=(\langle c|e\rangle|a\rangle)\otimes (\langle d|f\rangle|b\rangle)$$
and
$$(|a\rangle\langle c|\otimes|b\rangle\langle d|)|e,f\rangle\equiv (|a\rangle\langle c|\otimes|b\rangle\langle d|)(|e\rangle\otimes |f\rangle)\ =(|a\rangle\langle c|)|e\rangle\otimes (|b\rangle\langle d|)|f\rangle \equiv (\langle c|e\rangle|a\rangle)\otimes (\langle d|f\rangle|b\rangle)$$
So
$$(|a,b\rangle\langle c,d|)=(|a\rangle\langle c|\otimes|b\rangle\langle d|)$$

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